In Young's double slit experiment, two slits 0.12 mm apart are illuminated by monochromatic light of wavelength 480 nm. The screen is 1.0 m away from the slits. The distance of the second bright fringe from the central maximum will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 8\times10^{-3}\,\mathrm{m} \)
In Young's double slit experiment the \( n \)th bright fringe is at a distance \( y_n = n\lambda D/d \) from the central maximum. For the second bright fringe, \( n = 2 \).
\( d = 0.12\,\mathrm{mm} = 1.2\times10^{-4}\,\mathrm{m} \)
\( \lambda = 480\,\mathrm{nm} = 4.8\times10^{-7}\,\mathrm{m} \)
\( D = 1.0\,\mathrm{m} \)
\( \displaystyle y_2 = \frac{2\lambda D}{d} = \frac{2 \times 4.8\times10^{-7} \times 1.0}{1.2\times10^{-4}} \)
\( \displaystyle y_2 = 8\times10^{-3}\,\mathrm{m} \)